Problem: What is the slope of the line tangent to $f(x) = -x^{2}+3x+5$ at $x = -2$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x+\Delta x)^{2}+3(x+\Delta x)+5) - (-x^{2}+3x+5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-(x^{2}+2x \Delta x+\Delta x^{2})+3(x+\Delta x)+5) - (-x^{2}+3x+5)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-x^{2}-2(x \Delta x)-\Delta x^{2}+3x+3(\Delta x)+5+x^{2}-3x-5}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2(x \Delta x)-\Delta x^{2}+3(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -2x-\Delta x+3$ $ = -2x+3$ $ = (-2)(-2)+3$ $ = 7$